UnitExam1Dsolutions

UnitExam1Dsolutions - ID: D UCSD Physics 2B Summer Session...

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ID: D 4 UCSD Physics 2B Summer Session Unit Exam 1D Charges, Force & Fields Answer Section MULTIPLE CHOICE 1. ANS: C By definition, E = F q = 6.9 × 10 2 N 3.3 × 3 C = 2.1 × 5 N C TOP: ELECTRIC FIELD 2. ANS: A ρ = charge volume = Q 4 3 π R 3 = 3 4 6.60 × 4 C 3.14 () 3.10 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3 Ê Ë Á Á Á Á Á Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ = 5.29 C m 3 TOP: VOLUME CHARGE DENSITY 3. ANS: B For a sphere with radial symmetry, we can take the volume element to be the product or the surface of a spherical shell times its thickness dV = A surface × dr = 4 r 2 . We integrate the charge elements dq = r dV r as spherical shells since the density depends only on the radius r Q = r = r 4 r 2 = o r R 4 r 2 0 R = 4 πρ o R r 3 0 R = 4 o R R 4 4 = R 3 o = 1.31 × 2 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3 7.29 × 5 C m 3 Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ = 5.15 × 2 C TOP: VOLUME CHARGE DENSITY 4. ANS: A Fundamental properties of conductors : Electric field is zero inside Electric field is perpendicular at surface Excess charge must reside at surface only TOP: CONDUCTOR FIELD CONCEPT 5. ANS: D This is a conversion problem, so use the identity e + = 1.60 × 19 C to create "unit fraction" # e + = totalcharge charge/proton = 985 × 9 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1 e + 1.60 × C Ê Ë Á Á Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ ˜ ˜ = 6.16 × 12 TOP: ELEMENTARY CHARGE
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ID: D 2 6. ANS: C Φ E = E A = EA cos θ = 73.1 × 10 6 N / C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 26.2 m () 2 cos 61.1 ° = 2.43 × 2 N m 2 C TOP: Electric Flux 7. ANS: A
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UnitExam1Dsolutions - ID: D UCSD Physics 2B Summer Session...

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